Schrijf, gegeven een positief geheel getal, een functie die waar retourneert als het gegeven getal een palindroom is, en anders onwaar. 12321 is bijvoorbeeld een palindroom, maar 1451 is geen palindroom.
Aanbevolen oefening De som van de cijfers is Pallindroom of niet Probeer het!
Methode 1:
Laat het gegeven getal zijn op een . Een eenvoudige methode voor dit probleem is om eerst te beginnen omgekeerde cijfers van op een en vergelijk vervolgens het omgekeerde van op een met op een . Als beide hetzelfde zijn, retourneer dan waar, anders onwaar.
Hieronder volgt een interessante methode, geïnspireerd op methode #2 van dit na. Het idee is om er een kopie van te maken op een en de kopie recursief doorgeven door middel van referentie, en slagen op een op waarde. In de recursieve aanroepen, verdeel op een met 10 terwijl u naar beneden gaat in de recursieboom. Terwijl u omhoog gaat in de recursieboom, deelt u de kopie door 10. Wanneer ze elkaar ontmoeten in een functie waarvoor alle onderliggende oproepen voorbij zijn, wordt het laatste cijfer van op een is het 1e cijfer vanaf het begin en het laatste cijfer van de kopie is het 1e cijfer vanaf het einde.
C++
// A recursive C++ program to check> // whether a given number> // is palindrome or not> #include> using> namespace> std;> > // A function that returns true only> // if num contains one> // digit> int> oneDigit(>int> num)> {> > >// Comparison operation is faster> >// than division> >// operation. So using following> >// instead of 'return num> >// / 10 == 0;'> >return> (num>= 0 && aantal<10);> }> > // A recursive function to find> // out whether num is> // palindrome or not. Initially, dupNum> // contains address of> // a copy of num.> bool> isPalUtil(>int> num,>int>* dupNum)> {> > >// Base case (needed for recursion> >// termination): This> >// statement mainly compares the> >// first digit with the> >// last digit> >if> (oneDigit(num))> >return> (num == (*dupNum) % 10);> > >// This is the key line in this> >// method. Note that all> >// recursive calls have a separate> >// copy of num, but they> >// all share same copy of *dupNum.> >// We divide num while> >// moving up the recursion tree> >if> (!isPalUtil(num / 10, dupNum))> >return> false>;> > >// The following statements are> >// executed when we move up> >// the recursion call tree> >*dupNum /= 10;> > >// At this point, if num%10 contains> >// i'th digit from> >// beginning, then (*dupNum)%10> >// contains i'th digit> >// from end> >return> (num % 10 == (*dupNum) % 10);> }> > // The main function that uses> // recursive function> // isPalUtil() to find out whether> // num is palindrome or not> int> isPal(>int> num)> {> > >// Check if num is negative,> >// make it positive> >if> (num <0)> >num = -num;> > >// Create a separate copy of num,> >// so that modifications> >// made to address dupNum don't> >// change the input number.> >// *dupNum = num> >int>* dupNum =>new> int>(num);> > >return> isPalUtil(num, dupNum);> }> > // Driver program to test> // above functions> int> main()> {> >int> n = 12321;> >isPal(n) ? cout <<>'Yes
'>: cout <<>'No'> << endl;> > >n = 12;> >isPal(n) ? cout <<>'Yes
'>: cout <<>'No'> << endl;> > >n = 88;> >isPal(n) ? cout <<>'Yes
'>: cout <<>'No'> << endl;> > >n = 8999;> >isPal(n) ? cout <<>'Yes
'>: cout <<>'No'>;> >return> 0;> }> > // this code is contributed by shivanisinghss2110> |
>
>
C
#include> #include> > // A function that returns true only> // if num contains one digit> int> oneDigit(>int> num)> {> >// Comparison operation is faster> >// than division operation.> >// So using the following instead of 'return num / 10 == 0;'> >return> (num>= 0 && aantal<10);> }> > // A recursive function to find out whether> // num is palindrome or not.> // Initially, dupNum contains the address of a copy of num.> bool> isPalUtil(>int> num,>int>* dupNum)> {> >// Base case (needed for recursion termination):> >// This statement mainly compares the first digit with the last digit.> >if> (oneDigit(num))> >return> (num == (*dupNum) % 10);> > >// This is the key line in this method.> >// Note that all recursive calls have a separate copy of num,> >// but they all share the same copy of *dupNum.> >// We divide num while moving up the recursion tree.> >if> (!isPalUtil(num / 10, dupNum))> >return> false>;> > >// The following statements are executed when we move up the recursion call tree.> >*dupNum /= 10;> > >// At this point, if num % 10 contains the i'th digit from the beginning,> >// then (*dupNum) % 10 contains the i'th digit from the end.> >return> (num % 10 == (*dupNum) % 10);> }> > // The main function that uses the recursive function> // isPalUtil() to find out whether num is palindrome or not.> bool> isPal(>int> num)> {> >// Check if num is negative, make it positive.> >if> (num <0)> >num = -num;> > >// Create a separate copy of num, so that modifications> >// made to the address dupNum don't change the input number.> >int> dupNum = num;> > >return> isPalUtil(num, &dupNum);> }> > // Driver program to test above functions> int> main()> {> >int> n = 12321;> >isPal(n) ?>printf>(>'Yes
'>) :>printf>(>'No
'>);> > >n = 12;> >isPal(n) ?>printf>(>'Yes
'>) :>printf>(>'No
'>);> > >n = 88;> >isPal(n) ?>printf>(>'Yes
'>) :>printf>(>'No
'>);> > >n = 8999;> >isPal(n) ?>printf>(>'Yes
'>) :>printf>(>'No
'>);> > >return> 0;> }> |
>
>
Java
// A recursive Java program to> // check whether a given number> // is palindrome or not> import> java.io.*;> import> java.util.*;> > public> class> CheckPalindromeNumberRecursion {> > >// A function that returns true> >// only if num contains one digit> >public> static> int> oneDigit(>int> num) {> > >if> ((num>=>0>) && (num <>10>))> >return> 1>;> >else> >return> 0>;> >}> > >public> static> int> isPalUtil> >(>int> num,>int> dupNum)>throws> Exception {> > >// base condition to return once we> >// move past first digit> >if> (num ==>0>) {> >return> dupNum;> >}>else> {> >dupNum = isPalUtil(num />10>, dupNum);> >}> > >// Check for equality of first digit of> >// num and dupNum> >if> (num %>10> == dupNum %>10>) {> >// if first digit values of num and> >// dupNum are equal divide dupNum> >// value by 10 to keep moving in sync> >// with num.> >return> dupNum />10>;> >}>else> {> >// At position values are not> >// matching throw exception and exit.> >// no need to proceed further.> >throw> new> Exception();> >}> > >}> > >public> static> int> isPal(>int> num)> >throws> Exception {> > >if> (num <>0>)> >num = (-num);> > >int> dupNum = (num);> > >return> isPalUtil(num, dupNum);> >}> > >public> static> void> main(String args[]) {> > >int> n =>12421>;> >try> {> >isPal(n);> >System.out.println(>'Yes'>);> >}>catch> (Exception e) {> >System.out.println(>'No'>);> >}> >n =>1231>;> >try> {> >isPal(n);> >System.out.println(>'Yes'>);> >}>catch> (Exception e) {> >System.out.println(>'No'>);> >}> > >n =>12>;> >try> {> >isPal(n);> >System.out.println(>'Yes'>);> >}>catch> (Exception e) {> >System.out.println(>'No'>);> >}> > >n =>88>;> >try> {> >isPal(n);> >System.out.println(>'Yes'>);> >}>catch> (Exception e) {> >System.out.println(>'No'>);> >}> > >n =>8999>;> >try> {> >isPal(n);> >System.out.println(>'Yes'>);> >}>catch> (Exception e) {> >System.out.println(>'No'>);> >}> >}> }> > // This code is contributed> // by Nasir J> |
>
>
Python3
# A recursive Python3 program to check> # whether a given number is palindrome or not> > # A function that returns true> # only if num contains one digit> def> oneDigit(num):> > ># comparison operation is faster> ># than division operation. So> ># using following instead of> ># 'return num / 10 == 0;'> >return> ((num>>=> 0>)>and> >(num <>10>))> > # A recursive function to find> # out whether num is palindrome> # or not. Initially, dupNum> # contains address of a copy of num.> def> isPalUtil(num, dupNum):> > ># Base case (needed for recursion> ># termination): This statement> ># mainly compares the first digit> ># with the last digit> >if> oneDigit(num):> >return> (num>=>=> (dupNum[>0>])>%> 10>)> > ># This is the key line in this> ># method. Note that all recursive> ># calls have a separate copy of> ># num, but they all share same> ># copy of *dupNum. We divide num> ># while moving up the recursion tree> >if> not> isPalUtil(num>/>/>10>, dupNum):> >return> False> > ># The following statements are> ># executed when we move up the> ># recursion call tree> >dupNum[>0>]>=> dupNum[>0>]>/>/>10> > ># At this point, if num%10> ># contains i'th digit from> ># beginning, then (*dupNum)%10> ># contains i'th digit from end> >return> (num>%> 10> =>=> (dupNum[>0>])>%> 10>)> > # The main function that uses> # recursive function isPalUtil()> # to find out whether num is> # palindrome or not> def> isPal(num):> ># If num is negative,> ># make it positive> >if> (num <>0>):> >num>=> (>->num)> > ># Create a separate copy of> ># num, so that modifications> ># made to address dupNum> ># don't change the input number.> >dupNum>=> [num]># *dupNum = num> > >return> isPalUtil(num, dupNum)> > # Driver Code> n>=> 12321> if> isPal(n):> >print>(>'Yes'>)> else>:> >print>(>'No'>)> > n>=> 12> if> isPal(n) :> >print>(>'Yes'>)> else>:> >print>(>'No'>)> > n>=> 88> if> isPal(n) :> >print>(>'Yes'>)> else>:> >print>(>'No'>)> > n>=> 8999> if> isPal(n) :> >print>(>'Yes'>)> else>:> >print>(>'No'>)> > # This code is contributed by mits> |
>
>
C#
// A recursive C# program to> // check whether a given number> // is palindrome or not> using> System;> > class> GFG> {> > // A function that returns true> // only if num contains one digit> public> static> int> oneDigit(>int> num)> {> >// comparison operation is> >// faster than division> >// operation. So using> >// following instead of> >// 'return num / 10 == 0;'> >if>((num>= 0) &&(getal<10))> >return> 1;> >else> >return> 0;> }> > // A recursive function to> // find out whether num is> // palindrome or not.> // Initially, dupNum contains> // address of a copy of num.> public> static> int> isPalUtil(>int> num,> >int> dupNum)> {> >// Base case (needed for recursion> >// termination): This statement> >// mainly compares the first digit> >// with the last digit> >if> (oneDigit(num) == 1)> >if>(num == (dupNum) % 10)> >return> 1;> >else> >return> 0;> > >// This is the key line in> >// this method. Note that> >// all recursive calls have> >// a separate copy of num,> >// but they all share same> >// copy of *dupNum. We divide> >// num while moving up the> >// recursion tree> >if> (isPalUtil((>int>)(num / 10), dupNum) == 0)> >return> -1;> > >// The following statements> >// are executed when we move> >// up the recursion call tree> >dupNum = (>int>)(dupNum / 10);> > >// At this point, if num%10> >// contains i'th digit from> >// beginning, then (*dupNum)%10> >// contains i'th digit from end> >if>(num % 10 == (dupNum) % 10)> >return> 1;> >else> >return> 0;> }> > // The main function that uses> // recursive function isPalUtil()> // to find out whether num is> // palindrome or not> public> static> int> isPal(>int> num)> {> >// If num is negative,> >// make it positive> >if> (num <0)> >num = (-num);> > >// Create a separate copy> >// of num, so that modifications> >// made to address dupNum> >// don't change the input number.> >int> dupNum = (num);>// *dupNum = num> > >return> isPalUtil(num, dupNum);> }> > // Driver Code> public> static> void> Main()> {> int> n = 12321;> if>(isPal(n) == 0)> >Console.WriteLine(>'Yes'>);> else> >Console.WriteLine(>'No'>);> > n = 12;> if>(isPal(n) == 0)> >Console.WriteLine(>'Yes'>);> else> >Console.WriteLine(>'No'>);> > n = 88;> if>(isPal(n) == 1)> >Console.WriteLine(>'Yes'>);> else> >Console.WriteLine(>'No'>);> > n = 8999;> if>(isPal(n) == 0)> >Console.WriteLine(>'Yes'>);> else> >Console.WriteLine(>'No'>);> }> }> > // This code is contributed by mits> |
>
>
Javascript
> // A recursive javascript program to> // check whether a given number> // is palindrome or not> > >// A function that returns true> >// only if num contains one digit> >function> oneDigit(num) {> > >if> ((num>= 0) && (getal<10))> >return> 1;> >else> >return> 0;> >}> > >function> isPalUtil> >(num , dupNum) {> > >// base condition to return once we> >// move past first digit> >if> (num == 0) {> >return> dupNum;> >}>else> {> >dupNum = isPalUtil(parseInt(num / 10), dupNum);> >}> > >// Check for equality of first digit of> >// num and dupNum> >if> (num % 10 == dupNum % 10) {> >// if first digit values of num and> >// dupNum are equal divide dupNum> >// value by 10 to keep moving in sync> >// with num.> >return> parseInt(dupNum / 10);> >}>else> {> >// At position values are not> >// matching throw exception and exit.> >// no need to proceed further.> >throw> e;> >}> > >}> > >function> isPal(num)> >{> > >if> (num <0)> >num = (-num);> > >var> dupNum = (num);> > >return> isPalUtil(num, dupNum);> >}> > > > >var> n = 1242;> >try> {> >isPal(n);> >document.write(>' Yes'>);> >}>catch> (e) {> >document.write(>' No'>);> >}> >n = 1231;> >try> {> >isPal(n);> >document.write(>' Yes'>);> >}>catch> (e) {> >document.write(>' No'>);> >}> > >n = 12;> >try> {> >isPal(n);> >document.write(>' Yes'>);> >}>catch> (e) {> >document.write(>' No'>);> >}> > >n = 88;> >try> {> >isPal(n);> >document.write(>' Yes'>);> >}>catch> (e) {> >document.write(>' No'>);> >}> > >n = 8999;> >try> {> >isPal(n);> >document.write(>' Yes'>);> >}>catch> (e) {> >document.write(>' No'>);> >}> > // This code is contributed by Amit Katiyar> > |
>
>
PHP
// A recursive PHP program to // check whether a given number // is palindrome or not // A function that returns true // only if num contains one digit function oneDigit($num) { // comparison operation is faster // than division operation. So // using following instead of // 'return num / 10 == 0;' return (($num>= 0) && ($num<10)); } // A recursive function to find // out whether num is palindrome // or not. Initially, dupNum // contains address of a copy of num. function isPalUtil($num, $dupNum) { // Base case (needed for recursion // termination): This statement // mainly compares the first digit // with the last digit if (oneDigit($num)) return ($num == ($dupNum) % 10); // This is the key line in this // method. Note that all recursive // calls have a separate copy of // num, but they all share same // copy of *dupNum. We divide num // while moving up the recursion tree if (!isPalUtil((int)($num / 10), $dupNum)) return -1; // The following statements are // executed when we move up the // recursion call tree $dupNum = (int)($dupNum / 10); // At this point, if num%10 // contains i'th digit from // beginning, then (*dupNum)%10 // contains i'th digit from end return ($num % 10 == ($dupNum) % 10); } // The main function that uses // recursive function isPalUtil() // to find out whether num is // palindrome or not function isPal($num) { // If num is negative, // make it positive if ($num <0) $num = (-$num); // Create a separate copy of // num, so that modifications // made to address dupNum // don't change the input number. $dupNum = ($num); // *dupNum = num return isPalUtil($num, $dupNum); } // Driver Code $n = 12321; if(isPal($n) == 0) echo 'Yes
'; else echo 'No
'; $n = 12; if(isPal($n) == 0) echo 'Yes
'; else echo 'No
'; $n = 88; if(isPal($n) == 1) echo 'Yes
'; else echo 'No
'; $n = 8999; if(isPal($n) == 0) echo 'Yes
'; else echo 'No
'; // This code is contributed by m_kit ?>> |
>
>Uitvoer
Yes No Yes No>
Tijdcomplexiteit: O(logboek n)
Hulpruimte: O(logboek n)
Een getal controleren is palindroom of niet, zonder extra spatie te gebruiken
Methode 2: De methode string() gebruiken
- Wanneer het aantal cijfers van dat getal groter is dan 1018, kunnen we dat getal niet als een geheel getal beschouwen, omdat het bereik van long long int niet voldoet aan het gegeven getal.
- Dus neem de invoer als een string, voer een lus uit van begin tot lengte/2 en controleer het eerste teken (numeriek) tot het laatste teken van de string en de op één na laatste, enzovoort .... Als een teken niet overeenkomt, wordt de string zou geen palindroom zijn.
Hieronder vindt u de implementatie van bovenstaande aanpak
C++14
// C++ implementation of the above approach> #include> using> namespace> std;> > // Function to check palindrome> int> checkPalindrome(string str)> {> >// Calculating string length> >int> len = str.length();> > >// Traversing through the string> >// upto half its length> >for> (>int> i = 0; i // Comparing i th character // from starting and len-i // th character from end if (str[i] != str[len - i - 1]) return false; } // If the above loop doesn't return then it is // palindrome return true; } // Driver Code int main() { // taking number as string string st = '112233445566778899000000998877665544332211'; if (checkPalindrome(st) == true) cout << 'Yes'; else cout << 'No'; return 0; } // this code is written by vikkycirus> |
lang om Java te stringen
>
>
Java
// Java implementation of the above approach> import> java.io.*;> > class> GFG{> > // Function to check palindrome> static> boolean> checkPalindrome(String str)> {> > >// Calculating string length> >int> len = str.length();> > >// Traversing through the string> >// upto half its length> >for>(>int> i =>0>; i 2; i++) { // Comparing i th character // from starting and len-i // th character from end if (str.charAt(i) != str.charAt(len - i - 1)) return false; } // If the above loop doesn't return then // it is palindrome return true; } // Driver Code public static void main(String[] args) { // Taking number as string String st = '112233445566778899000000998877665544332211'; if (checkPalindrome(st) == true) System.out.print('Yes'); else System.out.print('No'); } } // This code is contributed by subhammahato348> |
>
>
Python3
# Python3 implementation of the above approach> > # function to check palindrome> def> checkPalindrome(>str>):> > ># Run loop from 0 to len/2> >for> i>in> range>(>0>,>len>(>str>)>/>/>2>):> >if> str>[i] !>=> str>[>len>(>str>)>->i>->1>]:> >return> False> > ># If the above loop doesn't> >#return then it is palindrome> >return> True> > > # Driver code> st>=> '112233445566778899000000998877665544332211'> if>(checkPalindrome(st)>=>=> True>):> >print>(>'it is a palindrome'>)> else>:> >print>(>'It is not a palindrome'>)> |
>
>
C#
// C# implementation of the above approach> using> System;> > class> GFG{> > // Function to check palindrome> static> bool> checkPalindrome(>string> str)> {> > >// Calculating string length> >int> len = str.Length;> > >// Traversing through the string> >// upto half its length> >for>(>int> i = 0; i { // Comparing i th character // from starting and len-i // th character from end if (str[i] != str[len - i - 1]) return false; } // If the above loop doesn't return then // it is palindrome return true; } // Driver Code public static void Main() { // Taking number as string string st = '112233445566778899000000998877665544332211'; if (checkPalindrome(st) == true) Console.Write('Yes'); else Console.Write('No'); } } // This code is contributed by subhammahato348> |
>
>
Javascript
Java-schakelaar
> > // Javascript implementation of the above approach> > // Function to check palindrome> function> checkPalindrome(str)> {> >// Calculating string length> >var> len = str.length;> > >// Traversing through the string> >// upto half its length> >for> (>var> i = 0; i // Comparing ith character // from starting and len-ith // character from end if (str[i] != str[len - i - 1]) return false; } // If the above loop doesn't return then it is // palindrome return true; } // Driver Code // taking number as string let st = '112233445566778899000000998877665544332211'; if (checkPalindrome(st) == true) document.write('Yes'); else document.write('No'); // This code is contributed by Mayank Tyagi> |
>
>Uitvoer
Yes>
Tijdcomplexiteit: O(|str|)
Hulpruimte : O(1)
Methode 3:
Hier is de eenvoudigste manier om te controleren of een getal Palindroom is of niet. Deze aanpak kan worden gebruikt als het aantal cijfers in het gegeven getal kleiner is dan 10^18, want als het aantal cijfers van dat getal groter is dan 10^18, kunnen we dat getal niet als een geheel getal beschouwen, aangezien het bereik van lang lang int voldoet niet aan het opgegeven getal.
Om te controleren of het gegeven getal een palindroom is of niet, draaien we gewoon de cijfers van het gegeven getal om en controleren of het omgekeerde van dat getal gelijk is aan het oorspronkelijke getal of niet. Als het omgekeerde van het getal gelijk is aan dat getal, dan zal het getal een Palindroom zijn, anders is het geen Palindroom.
C++
// C++ program to check if a number is Palindrome> #include> using> namespace> std;> // Function to check Palindrome> bool> checkPalindrome(>int> n)> {> >int> reverse = 0;> >int> temp = n;> >while> (temp != 0) {> >reverse = (reverse * 10) + (temp % 10);> >temp = temp / 10;> >}> >return> (reverse> >== n);>// if it is true then it will return 1;> >// else if false it will return 0;> }> int> main()> {> >int> n = 7007;> >if> (checkPalindrome(n) == 1) {> >cout <<>'Yes
'>;> >}> >else> {> >cout <<>'No
'>;> >}> >return> 0;> }> // This code is contributed by Suruchi Kumari> |
>
>
Java
/*package whatever //do not write package name here */> > import> java.io.*;> > class> GFG {> >// Java program to check if a number is Palindrome> > >// Function to check Palindrome> >static> boolean> checkPalindrome(>int> n)> >{> >int> reverse =>0>;> >int> temp = n;> >while> (temp !=>0>) {> >reverse = (reverse *>10>) + (temp %>10>);> >temp = temp />10>;> >}> >return> (reverse == n);>// if it is true then it will return 1;> >// else if false it will return 0;> >}> > >// Driver Code> >public> static> void> main(String args[])> >{> >int> n =>7007>;> >if> (checkPalindrome(n) ==>true>) {> >System.out.println(>'Yes'>);> >}> >else> {> >System.out.println(>'No'>);> >}> >}> }> > // This code is contributed by shinjanpatra> |
>
>
Python3
# Python3 program to check if a number is Palindrome> > # Function to check Palindrome> def> checkPalindrome(n):> > >reverse>=> 0> >temp>=> n> >while> (temp !>=> 0>):> >reverse>=> (reverse>*> 10>)>+> (temp>%> 10>)> >temp>=> temp>/>/> 10> > >return> (reverse>=>=> n)># if it is true then it will return 1;> ># else if false it will return 0;> > # driver code> n>=> 7007> if> (checkPalindrome(n)>=>=> 1>):> >print>(>'Yes'>)> > else>:> >print>(>'No'>)> > # This code is contributed by shinjanpatra> |
>
>
C#
// C# program to check if a number is Palindrome> > using> System;> > class> GFG {> > >// Function to check Palindrome> >static> bool> checkPalindrome(>int> n)> >{> >int> reverse = 0;> >int> temp = n;> >while> (temp != 0) {> >reverse = (reverse * 10) + (temp % 10);> >temp = temp / 10;> >}> >return> (> >reverse> >== n);>// if it is true then it will return 1;> >// else if false it will return 0;> >}> > >// Driver Code> >public> static> void> Main(>string>[] args)> >{> >int> n = 7007;> >if> (checkPalindrome(n) ==>true>) {> >Console.WriteLine(>'Yes'>);> >}> >else> {> >Console.WriteLine(>'No'>);> >}> >}> }> > // This code is contributed by phasing17> |
>
>
Javascript
> > // JavaScript program to check if a number is Palindrome> > // Function to check Palindrome> function> checkPalindrome(n)> {> >let reverse = 0;> >let temp = n;> >while> (temp != 0) {> >reverse = (reverse * 10) + (temp % 10);> >temp = Math.floor(temp / 10);> >}> >return> (reverse == n);>// if it is true then it will return 1;> >// else if false it will return 0;> }> > // driver code> > let n = 7007;> if> (checkPalindrome(n) == 1) {> >document.write(>'Yes'>,>''>);> }> else> {> >document.write(>'No'>,>''>);> }> > > // This code is contributed by shinjanpatra> > > |
>
>Uitvoer
Yes>
Tijdcomplexiteit: O(logboek10(n)) of O(aantal cijfers in een bepaald getal)
Hulpruimte : O(1) of constante
Dit artikel is samengesteld doorAashish Barnwal.